= Solution
At the hemispherical surface $r=a$, the outward normal from the conductor into the gap is $\widehat{\mathbf r}$. The tangential field is zero there and
$$
E_n=E_r(a,\theta)=-3E_0\cos\theta.
$$
The <electrostatic interface boundary conditions> give <surface charge density> $\sigma=\varepsilon_0E_n$, because the interior <electric field> is zero. Hence
$$
\boxed{\sigma_{\rm hemi}=-\frac{3\varepsilon_0V}{d}\cos\theta.}
$$
It is negative for a positive upper-plate voltage, greatest in magnitude at the apex, and zero at the equator. Its units are coulombs per square metre.
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